What are the solutions to the equation 2x+3=(x+1)^2 ?

The solutions to the equation 2x+3=(x+1)^2 are x=sqrt(2),x=-sqrt(2)

Find the zeros of 2x+3=(x+1)^2

The zeros of 2x+3=(x+1)^2 are x=sqrt(2),x=-sqrt(2)

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2 x + 3 = (x + 1)^{2}

x = 2, x = - 2

Decimal Notation

x = 1.41421 \dots, x = - 1.41421 \dots

Solution steps

2 x + 3 = (x + 1)^{2}

Expand

(x + 1)^{2} : x^{2} + 2 x + 1

2 x + 3 = x^{2} + 2 x + 1

Switch sides

x^{2} + 2 x + 1 = 2 x + 3

Move

1

to the right side

x^{2} + 2 x = 2 x + 2

Move

2 x

to the left side

x^{2} = 2

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 2, x = - 2

a^{2} + 1^{2} = 4^{2}

\frac{2}{3} x^{2} - 6 = 0

x, x^{2} + y^{2} = c^{2}

b x^{2} - a x = a + b

x (- 2 x - 11) = - 3

What are the solutions to the equation 2x+3=(x+1)^2 ?
The solutions to the equation 2x+3=(x+1)^2 are x=sqrt(2),x=-sqrt(2)
Find the zeros of 2x+3=(x+1)^2
The zeros of 2x+3=(x+1)^2 are x=sqrt(2),x=-sqrt(2)